jmc

Since $90^3=729,000$, $\text{AB}$ is greater than 90. Therefore, $\text{A}=9$. Since the ones digit of ${\text{AB}}^3$ is 3, $\text{AB}$ must be odd. The ones digit of ${\text{AB}}^3$ is the same as the ones digit of ${\text{B}}^3$, so we look at the ones digits of the cubes of the odd digits. $$\begin{array}{c}\text{The ones digit of }{1}^3\text{ is }1.\\ \text{The ones digit of }{3}^3\text{ is }7.\\ \text{The ones digit of }{5}^3\text{ is }5.\\ \text{The ones digit of }{7}^3\text{ is }3.\\ \text{The ones digit of }{9}^3\text{ is }9.\end{array}$$ Only $7^3$ has a ones digit of 3, so $\text{B}=7$. Therefore, $\text{A}+\text{B}=9+7=\boxed{16}$.